A submarine $A$ travelling at $17 \mathrm{~m} / \mathrm{s}$ is being chased along the line of its velocity by another submarine $B$ travelling at $34 \mathrm{~m} / \mathrm{s}$. B sends a sonar signal of $600 \mathrm{~Hz}$ to detect $A$ and receives a reflected sound of frequency $v$. The of $v$ is [Speed of sound in water $=1500 \mathrm{~ms}^{-1}$ ]

• Correct answer
  A

  $613.7 \mathrm{~Hz}$
• B

  $6137 \mathrm{~Hz}$
• C

  $62 \mathrm{~Hz}$
• D

  $539 \mathrm{~Hz}$

Solution

$\therefore$ Given, velocity of submarine $A, v_A=17 \mathrm{~m} / \mathrm{s}$ Velocity of submarine $B, v_B=34 \mathrm{~m} / \mathrm{s}$ Signal sent by submarine $B$ is detected by submarine $A$ can be shown as
Freqency of the signal, $f_0=600 \mathrm{~Hz}$ Speed of sound in water $v_s=1500 \mathrm{~ms}^{-1}$ Frequency received by submarine $A$ is $$ f_1=\left(\frac{v_{\mathrm{S}}-v_A}{v_{\mathrm{S}}-v_B}\right) f_0=\left[\frac{1500-17}{1500-34}\right] \times 600 \mathrm{~Hz} $$ $$ f_1=\frac{1483}{1466} \times 600 $$ Frequency received by submarine $B$ is $$ f_2=\left(\frac{v_{\mathrm{S}}+v_B}{v_s+v_A}\right) f_1 $$ Substituting given values and $f_1$ value from Eq. (i), we get $$ \begin{aligned} f_2 & =\left(\frac{1500+34}{1500+17}\right) \times\left[\frac{1483}{1466} \times 600\right] \\ f_2 & =1.0112 \times 1.0115 \times 600 \\ f_2 & =613.7 \mathrm{~Hz} \end{aligned} $$

Hence, the correct answer is option: 'A'.